Answer:
68% of jazz CDs play between 45 and 59 minutes.
Step-by-step explanation:
<u>The correct question is:</u> The playing time X of jazz CDs has the normal distribution with mean 52 and standard deviation 7; N(52, 7).
According to the 68-95-99.7 rule, what percentage of jazz CDs play between 45 and 59 minutes?
Let X = <u>playing time of jazz CDs</u>
SO, X ~ Normal(
)
The z-score probability distribution for the normal distribution is given by;
Z =
~ N(0,1)
Now, according to the 68-95-99.7 rule, it is stated that;
- 68% of the data values lie within one standard deviation points from the mean.
- 95% of the data values lie within two standard deviation points from the mean.
- 99.7% of the data values lie within three standard deviation points from the mean.
Here, we have to find the percentage of jazz CDs play between 45 and 59 minutes;
For 45 minutes, z-score is =
= -1
For 59 minutes, z-score is =
= 1
This means that our data values lie within 1 standard deviation points, so it is stated that 68% of jazz CDs play between 45 and 59 minutes.
With what there’s nothing
, my friend I think you forgot to post the picture
Answer:
Step-by-step explanation:
1 + 3 = 4
so find the point 1/4 of the way from (-1, 2) to (7, 8)
x coordinate
Δx = ¼(7 - (-1)) = 2
x = x₀ + Δx
x = -1 + 2 = 1
y coordinate
Δy = ¼(8 - 2) = 1½
y = y₀ + Δy
y = 2 + 1½ = 3½
(1, 3½)
1 meter = 1000 millimeters
so, 5 meters = 5*1000 = 5000 millimeters
and 6000 millimeters = 6000 / 1000 = 6 meters